2
$\begingroup$

Let $b: \mathbb R_+\times\mathbb R_+\times \mathcal P\to\mathbb R$ be Lipschitz, where $\mathcal P$ denotes the set of probability measures $\mu$ on $\mathbb R_+$ of finite first moment and is endowed with the Wasserstein metric of order $1$. Consider the equation

$$(1+\alpha)c(t,\mu, \alpha)=\int_{\mathbb R_+} \big(b(t,x,\mu)+c(t,\mu,\alpha)\big)^+\mu(dx) - (1-\alpha)\big(b(t,0,\mu) + c(t,\mu,\alpha)\big)^+,\quad\quad (\ast)$$

for all $t\in\mathbb R_+$, $\mu\in\mathcal P$ and $\alpha\in [0,1]$. My questions are as follows :

  1. Does $(\ast)$ admit a unique solution $c: \mathbb R_+\times \mathcal P\times [0,1]\to\mathbb R$?

  2. If so, could this solution be Lipschitz? If not, could this solution be Lipschitz w.r.t. $(x,\mu)$ and continuous w.r.t. $\alpha$?

PS : Some special cases have been studied, e.g. $b\ge 0$ or $b\equiv b(t,\mu)$, where we may derive the explicit expression for $c$. So my question is rather for the case where $b$ may change sign and depend on $x$.

$\endgroup$

1 Answer 1

3
$\begingroup$

If $b(\cdot,0,\cdot)\ge 0$, then $(\ast)$ admits a unique solution that is Lipschitz.

For any (fixed) $t\in\mathbb R_+$, $\mu\in\mathcal P$ and $\alpha\in [0,1]$, define the function $F\equiv F_{t,\mu,\alpha}:\mathbb R\to\mathbb R$ by

$$F(z):=(1+\alpha)z + (1-\alpha)\big(b(t,0,\mu)+z\big)^+ - \int_{\mathbb R_+}\big(b(t,x,\mu)+z\big)^+\mu(dx).$$

Then it is clear that $F(\pm\infty)=\pm\infty$. Under the assumption, one has for $z<-b(t,0,\mu)$

$$F(z)\le (1+\alpha) z<0$$

and for $z\ge -b(t,0,\mu)$

$$F(z)=2z + (1-\alpha)b(t,0,\mu) - \int_{\mathbb R_+}\big(b(t,x,\mu)+z\big)^+\mu(dx)$$

is strictly increasing on $[-b(t,0,\mu),+\infty)$. Therefore, $F$ has a unique root, denoted by $c(t,\mu,\alpha)$, i.e. $F\big(c(t,\mu,\alpha)\big)=0$. Namely, $(\ast)$ admits a unique solution $c(t,\mu,\alpha)$.

To show the Lipschitz continuity, it suffices to estimate $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|$, $|c(t,\mu',\alpha)-c(t,\mu,\alpha)|$ and $|c(t,\mu,\alpha')-c(t,\mu,\alpha)|$. The arguments are almost the same, so WLOG we only compute $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|$. Note that the solution $c(t,\mu,\alpha)\ge -b(t,0,\mu)$ and thus $(\ast)$ can be rewritten as

\begin{eqnarray} 2c(t,\mu,\alpha) &=& \int_{\mathbb R_+}\big(b(t,x,\mu)+c(t,\mu,\alpha)\big)^+\mu(dx)-(1-\alpha)b(t,0,\mu) \\ 2c(t',\mu,\alpha) &=& \int_{\mathbb R_+}\big(b(t',x,\mu)+c(t',\mu,\alpha)\big)^+\mu(dx)-(1-\alpha)b(t,0,\mu). \\ \end{eqnarray}

Making the difference of the above two equations, one has

\begin{eqnarray} 2|c(t',\mu,\alpha)-c(t,\mu,\alpha)| &\le & \int_{\mathbb R_+}\Big[C|t'-t|+\big|c(t',\mu,\alpha)-c(t,\mu,\alpha)\big|\Big]\mu(dx) \\ &= & C|t'-t|+\big|c(t',\mu,\alpha)-c(t,\mu,\alpha)\big|, \end{eqnarray} which yields $|c(t',\mu,\alpha)-c(t,\mu,\alpha)|\le C|t'-t|$, where $C$ denotes the Lipschitz constant of $b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.